## Re: [EMHL] From my geometry textbook (and School folklore)

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• Dear Nikos ... Yes, it s the same. ... That is, the circumcircle of ABC. C A P B B C Let P = BB / CC trABC ~ trAB C == angBAC =
Message 1 of 2 , Jul 14, 2001
Dear Nikos

[Antreas]:
>>Let ABC be a triangle and consider an arbitrary triangle AB'C' such that:
>>AB'C' ~ [=similar to] ABC [the vertex A is in common].
>>Find the locus of the intersection of the lines BB' and CC'.
>>(I. Ioannides: Mathematics of D' of Gymnasium. Vol II.
>>OEDB: Athens, 1968, p. 52, #66)

[Nikos]:
>At first glance, if the order of AB, AC is the
>same as the order of AB', AC',

Yes, it's the same.

[Nikos]:
>we can say that if s = AC/AB then
>a rotation around A by angle BAC moves B to B''
>and the homothetic of B'' with ratio s is the point C.
>The same synthesis of tranformations moves B' to C'
>and hence BB' and CC' form an angle equal to angle BAC.
>So the locus of the intersection of BB' and CC' is a
>circular arc (and the symmetric about BC) every point
>of which sees BC with a constant angle BAC.

That is, the circumcircle of ABC.

C'

A P
B'

B C

Let P = BB' /\ CC'

trABC ~ trAB'C' ==>

angBAC = angB'AC' ==> angBAB' = angCAC' (1)

and

AB / AC = AB' / AC' or AB / AB' = AC / AC' (2)

(1) and (2) ==> trABB' ~ trACC' ==> angAB'B = angAC'C ==>
quadAB'PC' is cyclic ==> angBPC = angB'AC' = angBAC ==>
The locus of P is the circumcircle of ABC.

Greetings

Antreas
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